I have a problem again with irrational equation.

I have to solve this equation:

$\displaystyle \sqrt[3]{x} + \sqrt[3]{{2x - 3}} = \sqrt[3]{{12(x - 1)}}$

I have solve it like this:

$\displaystyle \left( {\sqrt[3]{x} + \sqrt[3]{{2x - 3}}} \right)^3 = \left( {\sqrt[3]{{12(x - 1)}}} \right)^3$

$\displaystyle x + 3\sqrt[3]{x}\sqrt[3]{{2x - 3}}(\sqrt[3]{x} + \sqrt[3]{{2x - 3}}) + 2x - 3 = 12x - 12 $

$\displaystyle \sqrt[3]{x}\sqrt[3]{{2x - 3}}(\sqrt[3]{x} + \sqrt[3]{{2x - 3}}) = 3x - 3 $

$\displaystyle \sqrt[3]{x}\sqrt[3]{{2x - 3}}\sqrt[3]{{12(x - 1)}} = 3x - 3 $

$\displaystyle \sqrt[3]{{x(2x - 3)(12x - 12)}} = 3x - 3 $

$\displaystyle x(2x - 3)4(3x - 3) = (3x - 3)^3 $

$\displaystyle 4x(2x - 3) = (3x - 3)^2$

$\displaystyle 8x^2 - 12x = 9x^2 - 18x + 9 $

$\displaystyle x^2 - 6x + 9 = 0$

$\displaystyle (x - 3)^2 = 0 $

$\displaystyle x = 3$

This is one solution but there is another one that is not obvious: $\displaystyle x=1$.

Now, I have come to that solution by assuming that right side of equation is 0: $\displaystyle \sqrt[3]{{12(x - 1)}}=0$

Can someone explain me how to know exactly what are ALL solutions of irrational equations?